A black body is at a temperature of 2880 K. The energy of radiation emitted by this…
Physics · JEE Advanced · NTA Exams — Thermodynamics
A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien constant b = 2.88 × 106 nm K. Then
- U1 = 0
- U3 = 0
- U1 > U2
- U2 > U1.
Answer
(D) U 2 > U 1 .
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